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JEE MainMathematicsThree Dimensional Geometry

A point Q lies on the line L₁ : x - 2 2 = y - 1 1 = z + 1 2 . It is given that Q is equidistant from the planes P₁ : x + 2y - 2z = 0 and P₂ : 2x - 2y + z + 5 = 0 . If the x -coordinate of Q is strictly negative, then the perpendicular distance of the point Q from the line L₂ : x 1 = y - 3 2 = z + 5 2 is ________.

Correct answer

3

Step-by-step solution

Let the coordinates of the point Q on the line L₁ be (2t+2, t+1, 2t-1) . The perpendicular distance of Q from the plane P₁ is d₁ = |(2t+2) + 2(t+1) - 2(2t-1)| 1^2 + 2^2 + (-2)^2 = |2t+2+2t+2-4t+2| 3 = |6| 3 = 2 The perpendicular distance of Q from the plane P₂ is d₂ = |2(2t+2) - 2(t+1) + (2t-1) + 5| 2^2 + (-2)^2 + 1^2 = |4t+4-2t-2+2t-1+5| 3 = |4t+6| 3 Since Q is equidistant from P₁ and P₂ , we have d₁ = d₂ . |4t+6| 3 = 2 |4t+6| = 6 4t+6 = 6 or 4t+6 = -6 t = 0 or t = -3 If t = 0 , Q is (2, 1, -1) . Here the x -coord

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